With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} l₂: r = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} where \( \lambda \) and \( \mu \) are scalar parameters and \( p \) is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2015 - Paper 4

From the previous equations, substituting ( \lambda ) from (1) into (3):

$p + 5\mu + 3\left(\frac{3\mu + 3}{0}\right) = -2$

Finding values for ( \mu ) we eliminate values to find ( p ):

Ultimately, solving simplifies to:

$p = 15$.

To find the acute angle ( \theta ) between lines l₁ and l₂, we use the formula:

$\cos(\theta) = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}| |\mathbf{d_2}|}$

Where ( \mathbf{d_1} = \begin{pmatrix} 0 \ 1 \ -3 \end{pmatrix} ) and ( \mathbf{d_2} = \begin{pmatrix} 3 \ 4 \ -5 \end{pmatrix}$$.

Calculating the dot product:

$\mathbf{d_1} \cdot \mathbf{d_2} = 0 \cdot 3 + 1 \cdot 4 + (-3) \cdot (-5) = 0 + 4 + 15 = 19$

Finding magnitudes:

$|\mathbf{d_1}| = \sqrt{0^2 + 1^2 + (-3)^2} = \sqrt{10}$

$|\mathbf{d_2}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50}$

Then,

$\cos(\theta) = \frac{19}{\sqrt{10} \sqrt{50}}$

Thus, ( \theta = \cos^{-1}\left(\frac{19}{10\sqrt{5}}\right) \approx 81.32 \text{ degrees}$$.

Let point B lie on l₂ when ( \mu = 1 ):

Coordinates of B are:

$\begin{pmatrix} 8 + 3 \\ 5 + 4 \\ -2 - 5 \end{pmatrix} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix}$

The position vector of B relative to A is:

$\mathbf{AB} = \begin{pmatrix} 11 - 5 \\ 9 - \frac{1}{3} \\ -7 - 3 \end{pmatrix} = \begin{pmatrix} 6 \\ \frac{26}{3} \\ -10 \end{pmatrix}$

The direction vector of line l₁ is:

$\mathbf{d_1} = \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}$

The shortest distance is given by the formula:

$d = \frac{|\mathbf{AB} \cdot \mathbf{d_1}|}{|\mathbf{d_1}|}$

Computing yields:

$d = \frac{6}{\sqrt{10}} = 7.46 \text{ at 3 significant figures.}$